# I Proof of Stokes' theorem

1. Apr 8, 2017

### davidge

How would one prove the Stokes' theorem for general cases? Namely that $$\int_{\partial M} W = \int_M \partial W$$ where $M$ is the manifold.

2. Apr 8, 2017

### Orodruin

Staff Emeritus
The exterior derivative of the differential form $\omega$ is normally written $d\omega$.

There should be at least an outline of a proof in any textbook covering it.

3. Apr 8, 2017

### davidge

Ok. I have tried a derivation. Can this be considered valid?

$M = \int dM$ and (maybe) $\int_{a}^{b}dM \approx dM$ if $|b-a| <<1$
So, $M$ could be written as $$M = \sum_{n=1}^{\infty} n \times \lim_{(b\ -\ a) \longrightarrow 0} \int_{a}^{b}dM = \sum_{n=1}^{\infty} n \ dM$$
So, $$\int_M dw = \int_{\sum_{n=1}^{\infty} n \ dM} dw \ \text{,}$$ which we can split in $$\int_{dM} dw + \int_{dM} dw + \int_{dM} dw \ + \ ...$$
Now, deriving both sides of $\int_{\partial M} w = \int_M dw$ with respect to $M$ (and using the above), we get $w = \sum_{n=1}^{\infty}n \ dw$, which I guess is correct.

I know that this is not a derivation of the theorem, anyways maybe it is a proof that the theorem is correct.

Last edited: Apr 8, 2017
4. Apr 9, 2017

### Orodruin

Staff Emeritus
To be perfectly honest, it seems that you are not at all familiar with these concepts. Essentially nothing that you proposed above is a valid identity and your notation is severely scrambled.

5. Apr 9, 2017

### davidge

So are you saying that $w$ is not equal to an infinite sum of its differentials $dw$
And are you saying that a integral over an interval $M$ cannot be decomposed into integrals over infinitely many intervals between the end-points of $M$

6. Apr 9, 2017

### Orodruin

Staff Emeritus
No, I am saying that essentially everything you have written shows that you do not really understand what the integral of a differential form means.
This makes no sense whatsoever. You need to make up your mind whether $M$ is the manifold or whether it is a differential form. Regardless there is no integral over an interval from $a$ to $b$ and even in the simple case of a one-dimensional integral over a single variable $M$ that does not make sense. Also, an integral of a differential form is not a manifold nor a differential form, it is a number.

This is just delving deeper with more misconceptions.

Here it is not even clear what you consider $dM$ to be. $\partial M$ that appears in Stokes' theorem is the boundary of $M$, not a small part of $M$.

You cannot differentiate with respect to a manifold. What do you think the result would be?

To me it seems clear that you need to go back and study the basics of calculus on manifolds, check in detail what differential forms are and how their integrals are defined. You will not get the right result by just guessing.

7. Apr 9, 2017

### davidge

Oh, I thought it was a small part of $M$.
Yes, I know it. So a way of doing things right would be to map $M$ into an interval in $\mathbb{R}^m$ ($m$ is the dimension of $M$), through a function $\phi$, so that we would change all the $M$ in the integrals by $\phi (M)$? If so, would my indentities become valid?

8. Apr 9, 2017

### Orodruin

Staff Emeritus
No. I am sorry to be so blunt, but it seems you do not have the prerequisite understanding of differential forms and their integration needed for Stokes' theorem. Again, I suggest going back to the basics until you really understand them.

9. Apr 9, 2017

### davidge

Ok, so I'll read some texts on the subject before I come back here with another thread.

10. Apr 17, 2017

### mathwonk

as i recall, stokes' theorem is just the fundamental theorem of calculus, plus fubini's theorem. i suggest reading lang's analysis I, chapter XX; at least that's where i became happy with it.

well since you asked about general cases, then you augment this proof for a rectangle by a partition of unity argument to globalize it. but the rectangle case is really the main idea.

Last edited: Apr 17, 2017